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Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. Is it possible to write an algorithm with O(log n) runtime complexity? Examples: Input: values = [1,3,5,6], target = 7. Output: 4 Input: values = [1], target = 0. Output: 0
Yes, it is possible to write an algorithm with O(log n) runtime complexity to solve this problem. This can be accomplished using a Binary Search algorithm. The pseudocode to implement a binary search is as follows:
javascript Copy code
// With a sorted array and a target value
function binarySearch(values, target){
// Set left and right boundaries
let left = 0;
let right = values.length - 1;
// Iterate until left does not
// cross right
while(left <= right) {
// Find middle index
let mid = Math.floor((left + right) / 2);
// Check if target is equal to middle element
if(target === values[mid]){
return mid;
}
// Check if target is lower than mid element
if(target < values[mid]){
right = mid - 1;
}
// Check if target is greater than mid element
if(target > values[mid]){
left = mid + 1;
}
}
// Return position to insert:
return left;
}
Maintaining a sorted array allows the algorithm to skip elements and search faster, making it possible to achieve O(log n) runtime complexity. Learn more about the binary search algorithm here: https://www.geeksforgeeks.org/binary-search/
What is binary search algorithm
Binary search is an algorithm for finding an element within a sorted array. It works by dividing the array into two halves, checking if the element you are searching for is equal to, or greater than the middle element, and then repeating the process with the appropriate subarray.
In the worst case, the binary search algorithm requires O(log n) comparisons to find an element. Most programming languages provide efficient implementations of binary search in the standard library.
For example, in Python, the bisect.bisect() function provides a binary search of a sorted list.
Below is an example of binary search in Python:
python Copy code
def binary_search(arr, x):
n = len(arr)
low = 0
high = n - 1
while low < high:
mid = (low + high) // 2
if arr[mid] == x:
return mid
elif arr[mid] < x:
low = mid + 1
else:
high = mid - 1
return -1 #element not found
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